Tutorial 2
GEM2507
Physical Question from Everyday Life
Odor and Photosynthesis
Q1.1 Estimate the volume one single air molecule occupies(on
average) in the lecture room (let us assume the temperature is 24
oC).
Q1.2 From an independent estimate of the volume occupied by a
single air molecule in the lecture room by using the ideal gas law.
P  1Atm  1.013x105 Pa,
T  240 C  297K
PV  nRT
1 1
Volume of
= nRT (1m ol)(8.314Jm ol K )(297K )
2 3
1 mole of


2
.
44
x
10
m
5
P
1.013x10 Pa
air
Volume of 1
molecule of air
Volume of 1 mol of air
=
Number of molecules in 1 mol of air
2.44x102
 26
3
 20
3


4
.
053
x
10
m

4
.
053
x
10
cm
6.02x1023
Q1.3 Deduce an estimate for radius of an air molecule in the
lecture room.
 The radius of an air molecule can be deduced from the density of water
 Since the density of water is 1000 times the density of air, the volume of
space occupied by the air molecule must be 1000 times larger than the
volume of one water molecules. Assuming that the water molecules are
closely spaced, we obtain
 4/3 π r3 = 4.053x 10-20 cm3 /1000
 R = 3.04 x 10-8 cm
Q1.4 Estimate the average distance between air molecules in the
lecture room.
Q1.5 Compute the mean free path of the air molecules in the
lecture room
air molecule
Assumption: the air
molecules occupy
the room uniformly
Volume occupied
by 1 air molecule
d
d
l
d
Mean free path
l
1

2
4 2nr
4 2
1

1
10
3
.
04
x
10
4.05x1026

 2.47x108 m
Number of molecules per unit volume = 1/(4.05x10-20 cm3)
Q1.6 Use the Avogadro’s number and the definition of the mole to
determine the mass of the oxygen molecule
 Mass of one oxygen molecule
 = mass of one mole of oxygen/ Avogadro’s number
 = 32 x 10-3 kg/6.02x1023 = 5.32 x 10-26 kg
Q1.7 Repeat the previous question for the nitrogen molecule and
then estimate the mean mass of an air molecule
Mass of one nitrogen molecule
= mass of one mole of nitrogen/ Avogadro’s number
= 28 x 10-3 kg/6.02x1023 = 4.65 x 10-26 kg
Mean mass of one air molecule
= (0.2 x 32 + 0.8 x 28) x 10-3 kg/ 6.02 x 1023
=4.78 x 10-26 kg
Q1.8 Estimate the average velocity of an air molecule in the
lecture room.
 ½ mv2 = 3/2 kbT
 ½ (4.78 x 10-26) v2 = 3/2 (1.38x10-23) x 298
 v = 508 m/s
Q1.9 Using the results of the previous questions, determine the
mean time between collisions of air molecules.
l
2.47x108 m


 4.86x1011 s
v
508m / s
Q1.10 Estimate the diffusion constant of the air molecules
1
1
D  l v  2.47 x10 8 x508  1.71x10 6 m 2 / s
3
3
Q1.11 Discuss how the results above would change if the
temperature was varied while keeping the pressure and volume in
the room constant.
 The velocity of air molecules will increase.
 Hence, there will be a decrease in the mean free time and increase in
diffusion constant
Q2. Name two key advantages of olfaction with regards to vision?
 Permanence
 Odors is more permanent than light in the sense




that a scent will remain for some time even after
the scent source has moved away while object can
no longer be seen when it has been moved away
Manufacturability
 Most organisms produce odor but not light
because many odors are simple organic
compounds.
Detection
 The spatial resolution of detected light is more
complicated hence the development of adequate
visual system is more complex than that of
olfactory system.
Odor can be useful in places where light cannot pass
such as muddy water, soil etc
Odor detection works as well in the dark as it does in
the light
Q3. We have discussed odors that spread through air. Could they
spread through water? What does this mean for fish?
 Yes, odors spread through water. The diffusion of odors in water and in
fact in any medium occurs through brownian motion (random collision
between the odorant and the medium particles. Since diffusion is
possible in water, olfaction sensor is quite highly developed for some
fishes for example shark which can smell as little as one part per
million of blood in water
Q4. On an inhabitable planet orbiting a significantly hotter star
than our sun, would the leaves more likely be blue or red?
 A hotter star will have a spectrum with an intensity peaks shifted toward
shorter wavelengths. Hence, we should expect that the leaves would
more likely be blue. Whether or not a leaf would really be blue depends
on many factors which likely cannot be answered until we discover
other planets with life.
Q5. How many photons does it take to split one water molecule?
Since two electrons need to be removed from two hydrogen atoms in water
then only two photons are necessary namely one for each electron harvested
. According to this website http://www.biologie.uni-hamburg.de/bonline/e24/24c.htm , there is only one photon needed to split two water
molecules
Q6. What would the color be of fluorescent chlorophyll
molecules?
 In fluorescence, light is absorbed and then later
emitted. In most cases, the fluorescent light is
only emitted when the molecule returns from the
first (lowest) excited state to the ground state. If it
was excited to a higher level than first excited
state, usually, the molecule first loses some energy
due e.g. rotation or vibration bringing it to the
first excited state.
Fluorescent
chlorophyll
 In the case of chlorophyll, there are two clear
peaks in the absorption. The one corresponding to
the longer wavelength, i.e. the peak around 650
nm (red light) is associated with the lowest
excited state and therefore gives the color of
fluorescent chlrocphyll
 In short: fluorescent chlorophyll is red
jellyfish
Q7. If an atom would have three different excited
levels, how many different wavelengths can the
absorbed and emitted photons have?
T23
 There are 6 possible wavelengths as shown in the
figure
T12
Ground state
Q8. Why are there two different lines in Figure 12.10?
 Chlorophyll has a number of
(closely related) different
structures. Each structure has a
slightly different absorption
spectrum. Most terrestial plants
have chlorophyll a
Q9. Is it conceivable to have a photosynthesis-like process in
leaves that are transparent of visible light (that is, the visible light
cannot be used as an energy source)?
 Yes, it is feasible. Whether it would be biologically useful is another
matter though. It may be reasonable to speculate that the wavelength
should not be too far from visible range because a too long wavelength
may not have enough energy to do something practically useful and on
the other hand the energy of a too short wavelength may be so large
that it destroys the photosynthesizing molecules. Moreover, the
quantity of photons is also important to light. Hence, there is an
evolutionary benefit in having photosynthesis using visible light since
the maximum intensity of sunlight is in the visible range.